Acil çözülmesi gereken bir SORU VAR.

1) In a beer factory, the waste water is being heated by a heat exchanger. The temperature of the heating water is 45 ᵒC and its flow rate is 25 m3/h. The inlet temperature of waste water recorded as 10 ˚C and its flow rate is 30 m3/h.


a) Calculate K and τ values for this heating system.
b) If the temperature of heating water is increased to 55 ᵒC at t=0, what will be the response equation of the output variable, y(t)=?
c) What will be outlet temperature of waste water at 5. hour?


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To calculate the K and τ values for the heating system, we can use the heat exchanger equation:
Q = K * (Ti - To) * e^(-t / τ)
where: Q is the heat transfer rate (in this case, it's the product of the mass flow rate and specific heat capacity), K is the system gain, Ti is the initial temperature of the heating water, To is the initial temperature of the waste water, t is the time, and τ is the time constant.
Given values: Ti = 45 °C To = 10 °C Flow rate of heating water (m1) = 25 m³/h Flow rate of waste water (m2) = 30 m³/h
First, let's calculate the heat transfer rate Q. We can assume that the specific heat capacity of water is 4.18 kJ/kg·°C.
Q = m2 * c * (To - Ti)
where c is the specific heat capacity of water.
Q = 30 * 10³ * 4.18 * (10 - 45) = -94.35 * 10⁶ kJ/h
a) Calculating K: K = Q / (m1 * (Ti - To))
K = (-94.35 * 10⁶) / (25 * (45 - 10)) = -4.19 * 10⁶ kJ/h·m³
b) Calculating τ: To find the value of τ, we need to determine the time it takes for the temperature of the waste water to reach 63.2% of its final value when the heating water temperature changes from 45 °C to 55 °C.
Using the exponential equation:
e^(-t / τ) = 0.632
Solving for t:

• t / τ = ln(0.632)

t = - τ * ln(0.632)
We know that at t = 0, the temperature of the heating water changes to 55 °C.
Thus, the response equation for the output variable, y(t), is given by:
y(t) = To + (Ti - To) * (1 - e^(-t / τ))
At t = 0, Ti = 55 °C and To = 10 °C.
c) To find the outlet temperature of the waste water at 5 hours, we can substitute t = 5 into the response equation:
y(t = 5) = To + (Ti - To) * (1 - e^(-5 / τ))

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ChatGPT sağ olsun

"To calculate the values of K and τ for the heating system, we can use the heat exchanger equation:
Q = K * (ΔT) Q = m * Cp * (T2 - T1)
Where: Q is the heat transfer rate (in kW) K is the heat transfer coefficient (in kW/°C) ΔT is the temperature difference (in °C) m is the mass flow rate (in kg/h) Cp is the specific heat capacity (in kW/kg·°C) T2 and T1 are the outlet and inlet temperatures, respectively (in °C)
First, let's convert the flow rates from m^3/h to kg/h using the density of water, which is approximately 1000 kg/m^3.
Mass flow rate of heating water (m1) = 25 m^3/h * 1000 kg/m^3 = 25000 kg/h Mass flow rate of waste water (m2) = 30 m^3/h * 1000 kg/m^3 = 30000 kg/h
a) Calculate K and τ values: ΔT = T1 - T2 = 10 ˚C - 45 ˚C = -35 ˚C
Let's assume a specific heat capacity of 4.18 kJ/kg·°C for water.
Q = K * (ΔT) Q = m * Cp * (T2 - T1)
25000 * 4.18 * (-35) = K * (-35) -3665000 = -35K K ≈ 104,714.29 kW/°C
τ = (m2 * Cp) / K τ = (30000 * 4.18) / 104714.29 τ ≈ 1.1981 hours
Therefore, the values of K and τ for this heating system are approximately K = 104,714.29 kW/°C and τ = 1.1981 hours.
b) If the temperature of the heating water is increased to 55 ˚C at t = 0, the response equation of the output variable, y(t), can be described by a first-order system response:
y(t) = y_ss + (y_0 - y_ss) * (1 - e^(-t/τ))
Where: y(t) is the output variable at time t y_ss is the steady-state value of the output variable y_0 is the initial value of the output variable at t = 0 τ is the time constant
In this case, the initial value of the output variable, y_0, is 10 ˚C (the initial temperature of the waste water) and the steady-state value, y_ss, is the outlet temperature of the waste water when the system reaches equilibrium.
To calculate y_ss, we can set t = ∞ in the response equation:
y_ss = y_0 + (y_ss - y_0) * (1 - e^(-∞/τ)) y_ss = y_0 + (y_ss - y_0) * (1 - 0) y_ss = y_ss
This means the steady-state value, y_ss, is equal to the initial temperature of the waste water, which is 10 ˚C.
Therefore, the response equation of the output variable, y(t), is:
y(t) = 10 ˚C + (55 ˚C - 10 ˚C) * (1 - e^(-t/1.1981))
c) To calculate the outlet temperature of the waste water at 5 hours, we can use the response equation with t = 5:
y(5) = 10 ˚C + (55 ˚C - 10 ˚C) * (1 - e^(-5/1.1981))
Calculating this equation will give you the outlet temperature of the waste water at 5 hours."

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